We have that the image distance of the object (obeying signs) is [tex]q=-7.61cm[/tex]
From the question we are told that:
Object distance [tex]d=14.5[/tex]
Power [tex]P=-6.30[/tex]
Object Height [tex]h=1.00mm=>0.1cm[/tex]
Generally, the equation for Lens Power is mathematically given by
[tex]P=\frac{1}{f}[/tex]
[tex]-6.3=\frac{1}{f}[/tex]
[tex]f=16cm[/tex]
Therefore
Apply Lens equation we have
[tex]\frac{1}{f}=\frac{1}{d}+\frac{1}{q}[/tex]
[tex]\frac{1}{q}=-\frac{1}{14.5}+\frac{1}{16}[/tex]
[tex]q=-7.61cm[/tex]
In conclusion
The image distance of the object (obeying signs) is [tex]q=-7.61cm[/tex]
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